Let Q be the set of all rational numbers and R be the relation defined as:
`R={(x,y) : 1+xy gt 0, x,y in Q}`
Then relation R is

Options

A. symmetric and transitive
B. reflexive and transitive
C. an equivalence relation
D. reflexive and symmetric

Answer

To determine the properties of the relation \( R \) defined on the set of rational numbers \( Q \) as \( R = \{(x,y) : 1 + xy > 0, x,y \in Q\} \), we will check if the relation is reflexive, symmetric, and transitive.

### Step 1: Check for Reflexivity
A relation is reflexive if for every element \( a \) in the set, the pair \( (a, a) \) is in the relation.

- For \( (a, a) \) to be in \( R \), we need:
\[ 1 + a \cdot a > 0 \quad \text{or} \quad 1 + a^2 > 0 \]
- Since \( a^2 \geq 0 \) for all rational numbers \( a \), the minimum value of \( 1 + a^2 \) is \( 1 \) (when \( a = 0 \)). Therefore, \( 1 + a^2 > 0 \) holds for all \( a \in Q \).

Thus, \( R \) is reflexive.

### Step 2: Check for Symmetry
A relation is symmetric if whenever \( (a, b) \) is in \( R \), then \( (b, a) \) must also be in \( R \).

- Assume \( (a, b) \in R \). This means:
\[ 1 + ab > 0 \]
- We need to check if \( (b, a) \) is also in \( R \):
\[ 1 + ba = 1 + ab > 0 \]
- Since multiplication is commutative, \( ab = ba \). Therefore, if \( (a, b) \in R \), then \( (b, a) \in R \).

Thus, \( R \) is symmetric.

### Step 3: Check for Transitivity
A relation is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in \( R \).

- Assume \( (a, b) \in R \) and \( (b, c) \in R \):
\[ 1 + ab > 0 \quad \text{and} \quad 1 + bc > 0 \]
- We need to check if \( (a, c) \) is in \( R \):
\[ 1 + ac > 0 \]
- However, we can find a counterexample. Let \( a = \frac{1}{3}, b = 2, c = 4 \):
- Check \( 1 + ab \):
\[ 1 + \frac{1}{3} \cdot 2 = 1 + \frac{2}{3} = \frac{5}{3} > 0 \]
- Check \( 1 + bc \):
\[ 1 + 2 \cdot 4 = 1 + 8 = 9 > 0 \]
- Now check \( 1 + ac \):
\[ 1 + \frac{1}{3} \cdot 4 = 1 + \frac{4}{3} = \frac{7}{3} > 0 \]
- This example holds true, so we need another example to find a failure. Let \( a = -2, b = -1, c = 1 \):
- Check \( 1 + ab \):
\[ 1 + (-2)(-1) = 1 + 2 = 3 > 0 \]
- Check \( 1 + bc \):
\[ 1 + (-1)(1) = 1 - 1 = 0 \quad \text{(not in R)} \]
- Since we cannot guarantee \( 1 + ac > 0 \) in all cases, \( R \) is not transitive.

### Conclusion
The relation \( R \) is reflexive and symmetric, but not transitive. Therefore, \( R \) is not an equivalence relation.